Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $y = \dfrac{-2a - 8}{a^2 + 2a - 35} \div \dfrac{a - 1}{3a^2 - 18a + 15} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{-2a - 8}{a^2 + 2a - 35} \times \dfrac{3a^2 - 18a + 15}{a - 1} $ First factor out any common factors. $y = \dfrac{-2(a + 4)}{a^2 + 2a - 35} \times \dfrac{3(a^2 - 6a + 5)}{a - 1} $ Then factor the quadratic expressions. $y = \dfrac {-2(a + 4)} {(a - 5)(a + 7)} \times \dfrac {3(a - 5)(a - 1)} {a - 1} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac {-2(a + 4) \times 3(a - 5)(a - 1) } { (a - 5)(a + 7) \times (a - 1)} $ $y = \dfrac {-6(a - 5)(a - 1)(a + 4)} {(a - 5)(a + 7)(a - 1)} $ Notice that $(a - 5)$ and $(a - 1)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {-6\cancel{(a - 5)}(a - 1)(a + 4)} {\cancel{(a - 5)}(a + 7)(a - 1)} $ We are dividing by $a - 5$ , so $a - 5 \neq 0$ Therefore, $a \neq 5$ $y = \dfrac {-6\cancel{(a - 5)}\cancel{(a - 1)}(a + 4)} {\cancel{(a - 5)}(a + 7)\cancel{(a - 1)}} $ We are dividing by $a - 1$ , so $a - 1 \neq 0$ Therefore, $a \neq 1$ $y = \dfrac {-6(a + 4)} {a + 7} $ $ y = \dfrac{-6(a + 4)}{a + 7}; a \neq 5; a \neq 1 $